CC BY
4URAL MATHEMATICAL JOURNAL, Vol. 7, No. 2, 2021, pp. 51-58
DOI: 10.15826/umj.2021.2.004
SHILLA GRAPHS WITH b = 5 AND b = 61
Alexander A. Makhnev^, Ivan N. Belousov^
Krasovskii Institute of Mathematics and Mechanics,
Ural Branch of the Russian Academy of Sciences, 16 S. Kovalevskaya Str., Ekaterinburg, 620990, Russia
Ural Federal University, 19 Mira str., Ekaterinburg, 620002, Russia
[email protected], [email protected]
Abstract: A Q-polynomial Shilla graph with b =5 has intersection arrays {105t, 4(21t + 1), 16(t + 1); 1, 4(t + 1), 84t}, t € {3, 4, 19}. The paper proves that distance-regular graphs with these intersection arrays do not exist. Moreover, feasible intersection arrays of Q-polynomial Shilla graphs with b = 6 are found.
Keywords: Shilla graph, Distance-regular graph, Q-polynomial graph.
1. Introduction
We consider undirected graphs without loops or multiple edges. For a vertex a of a graph r, denote by r(a) the ith neighborhood of a, i.e., the subgraph induced by r on the set of all vertices at distance i from a. Define [a] = r1(a) and a± = {a} U [a].
Let r be a graph, and let a, b € r. Denote by ^(a, b) (by A(a, b)) the number of vertices in [a] n [b] if a and b are at distance 2 (are adjacent) in r. Further, the induced [a] n [b] subgraph is called subgraph (A-subgraph).
If vertices u and w are at distance i in r, then we denote by bj(u, w) (by Cj(u, w)) the number of vertices in the intersection of Ti+1(u) (of ri-1(u), respectively) with [w]. A graph r of diameter d is called distance-regular with intersection array {b0, b1,..., bd-1; c1,..., cd} if, for each i = 0,..., d, the values b»(u, w) and c (u, w) are independent of the choice of vertices u and w at distance i in r. Define a» = k — b — c». Note that, for a distance regular graph, b0 is the degree of the graph and a1 is the degree of the local subgraph (the neighborhood of the vertex). Further, for vertices x and y at distance l in the graph r, denote by pj (x,y) the number of vertices in the subgraph r»(x) n Tj(y). The numbers pj(x,y) are called the intersection numbers of r (see [2]). In a distance-regular graph, they are independent of the choice of x and y.
A Shilla graph is a distance-regular graph r of diameter 3 with second eigenvalue equal to a = a3. In this case, a divides k and b is defined by b = b(r) = k/a. Morover, a1 = a — b and r has intersection array {ab, (a + 1)(b — 1), b2; 1, c2, a(b — 1)}. Feasible intersection arrays of Shilla graphs are found in [6] for b € {2, 3}.
Feasible intersection arrays of Shilla graphs are found in [1] for b = 4 (50 arrays) and for b = 5 (82 arrays). At present, a list of feasible intersection arrays of Shilla graphs for b = 6 is unknown. Moreover, the existence of Q-polynomial Shilla graphs with b = 5 also is unknown.
In this paper, we find feasible intersection arrays of Q-polynomial Shilla graphs with b = 6 and prove that Q-polynomial Shilla graphs with b = 5 do not exist.
1This work was supported by RFBR and NSFC (project № 20-51-53013).
Theorem 1. A Q-polynomial Shilla graph with b = 6 has intersection array
(1) {42t, 5(7t + 1), 3(t + 3); 1,3(t + 3), 35t}, where t € {7,12,17,27, 57};
(2) {372, 315, 75;1,15, 310}, {744, 625,125;1, 25, 620} or {930, 780,150; 1, 30, 775};
(3) {312,265,48;1,24, 260}, {624,525,80;1, 40, 520}, {1794,1500, 200; 1,100,1495} or {5694,4750,600;1, 300, 4745}.
In view of Theorem 2 from [1], a Q-polynomial Shilla graph with b = 5 has intersection array {105t, 4(21t + 1), 16(t + 1); 1, 4(t + 1), 84t}, t € {3, 4,19}.
Theorem 2. Distance-regular graphs with intersection arrays {315,256,64; 1,16, 252} and {1995, 1600, 320; 1, 80, 1596} do not exist.
Theorem 3. Distance-regular graphs with intersection array {420,340,80; 1,20,336} do not exist.
2. Proof of Theorem 1
In this section, r is a Q-polynomial Shilla graph with b = 6. Then (a2 — 5a — 6)2 — 4(5b2 — a2) is the square of an integer. By [6, Lemma 8], we have
2a < C2b(b + 1) + b2 — b — 2;
therefore, a < 21c2 +14. It follows from the proof of Theorem 9 in [6] that either k < b3 — b = 6 ■ 35 or v < k(2b3 — b + 1) = 428k. By [6, Corollary 17 and Theorem 20], the number b2 + c2 divides b(b — 1)b2 and
—34 = —b2 + 2 < 6>3 < —b2(b + 3)/(3b + 1) < —18. Theorem 2 from [7] implies the following lemma.
Lemma 1. If b2 = c2, then r has an intersection arrays {42t, 5(7t + 1), 3(t + 3);1, 3(t + 3), 35t} and t € {7,12,17,27, 57}.
To the end of this section, assume that b2 = c2 and k > > 02 > 03 are eigenvalues of the graph r. Then
6(6b2 + C2)/(b2 + C2) = —03.
On the other hand, according to [6, Lemma 10], the number c2 divides (a + 6)b2, 30a(a + 1) and (a + 6)b2 > (a + 1)c2.
Lemma 2. If —34 < 03 < —18, then one of the following statements holds:
(1) 03 = —31 and r has one of the intersection arrays {372,315,75; 1,15,310}, {744, 625,125; 1, 25, 620}, and {930, 780,150;1, 30, 775};
(2) 03 = —26 and r has one of the intersection arrays {312,265,48; 1,24,260}, {624, 525, 80; 1, 40, 520}, {1794,1500, 200; 1,100,1495}, and {5694, 4750, 600; 1, 300, 4745};
(3) 03 = —21 and r has one of the intersection arrays {42t, 5(7t + 1), 3(t + 3);1,3(t + 3), 35t} for t € {7,12,17, 27, 57}.
Proof. By [6, Lemma 10], c2 divides 6(6 — 1)62 = 3062 and, by [6, Corollary 17], the smallest nonprinciple eigenvalue 0з is equal to 6(662 + c2)/(62 + c2). Therefore, 30(0з + 6)/(0з + 36) is an integer and 6>з € {—34, —33, —32, —31, —30, —27, —26, —24, —21, —18}.
Let 03 = —34. Then 3(662 + c2) = 17(62 + c2) and 62 = 14c2. Further, 0з is a root of the equation x2 — (ai + a2 — k)x + (6 — 1)62 — a2 = 0; therefore, a = 425/28 ■ c2 — 34. In this case, the multiplicity of the first nonprincipal eigenvalue is m1 = 6/5 ■ (2545c2 — 5544)/c2, a contradiction with the fact that 5 does not divide 6 ■ 5544.
Let 6>з = —33. Then 2(662 + C2) = 11(6*2 + C2) and 62 = 9c2. Further, a = 275/27 ■ C2 — 33 and the multiplicity of the first nonprincipal eigenvalue is equal to m1 = 6/5 ■ (1645c2 — 5184)/c2, a contradiction as above.
Let 6>з = —32. Then 3(662 + C2) = 16(62 + C2) and 262 = 13c2. Further, a = 100/13 ■ C2 — 32 and the multiplicity of the first nonprincipal eigenvalue is m1 = 6/5 ■ (1195c2 — 4836)/c2, a contradiction as above.
Let 6>з = —31. Then 6(662 + C2) = 31(62 + C2) and 62 = 5c2. Further, a = 31/5 ■ C2 — 31 and the multiplicity of the first nonprincipal eigenvalue is m1 = 30(37c2 — 180)/c2 = 1110 — 5400/c2. The number of vertices in the graph is 31/5 ■ (222c2 — 2005c2 + 4500)/c2 ; hence, c2 divides 900 and is a multiple of 5. By computer enumeration, we find that, only for c2 = 15,25 and 30, we have admissible intersection arrays {372,315, 75; 1,15,310}, {744,625,125; 1,25,620} and {930,780,150; 1, 30, 775}.
Let 6>з = —30. Then (662 + C2) = 5(62 + C2) and 62 = 4c2. Further, a = 125/24 ■ C2 — 30 and the multiplicity of the first nonprincipal eigenvalue is m1 =6/5 ■ (745c2 — 4176)/c2, a contradiction as above.
Let 6>з = —27. Then 2(662 + C2) = 9(62 + C2) and 362 = 7c2. Further, a = 25/7 ■ C2 — 25 and the multiplicity of the first nonprincipal eigenvalue is m1 =6/5 ■ (445c2 — 3276)/c2, a contradiction as above.
Let 6>з = —26. Then 3(662 + C2) = 13(62 + C2) and 62 = 2c2. Further, a = 13/4 ■ C2 — 26 and the multiplicity of the first nonprincipal eigenvalue is m1 = 6(77c2 — 600)/c2 = 462 — 3600/c2. The number of vertices in the graph is 13/8 ■ (231c2 — 3340c2 + 12000)/c2; hence, c2 divides 1200 and is a multiple of 4. By computer enumeration, we find that only for c2 = 24,40,100, and 300 we have admissible intersection arrays {312,265,48; 1,24,260}, {624, 525,80; 1,40, 520}, {1794,1500,200;1,100,1495}, and {5694,4750,600; 1, 300,4745}.
Let 0з = —21. Then 2(662 + c2) = 7(62 + c2) and 62 = c2. Further, a = 7/3 ■ c2 — 21 and the multiplicity of the first nonprincipal eigenvalue is m1 = 6(41c2 — 360)/c2 = 246 — 2160/c2. The number of vertices in the graph is 7/3 ■ (82c2 — 1335c2 + 5400)/c2 ; hence, c2 divides 1080 and is a multiple of 3. By computer enumeration, we find that, only for c2 = 18,30,45,60,90, and 180, we have admissible intersection arrays {42t, 5(7t+1), 3(t+3); 1,3(t+3), 35t} for t € {3, 7,12,17,27, 57}. A graph with the array obtained for t = 3 does not exist by [5].
Let 6>з = —18. Then 6(662 + C2) = 19(62 + C2), so 362 = 2c2. Further, a = 2512 ■ C2 — 18 and the multiplicity of the first nonprincipal eigenvalue is m1 = 6/5 ■ (145c2 — 1224)/c2, a contradiction. The lemma is proved. □
Theorem 1 follows from Lemmas 1-2.
3. Triple intersection numbers
In the proof of Theorem 3, the triple intersection numbers [3] are used.
Let r be a distance-regular graph of diameter d. If u^u2,u3 are vertices of the graph r, then rj., r2, r3 are non-negative integers not greater than d. Denote by \ U^U3 ' the set of vertices
w € r such that d(w, u) = r and by
n\U2U'i
rir2r3
the number of vertices in
«1«2 «3 rir2r3
The numbers
«1«2«3 rir2r3
of
«1 «2«3 rir2r3
[rir2r3]. H
are called the triple intersection numbers. For a fixed triple of vertices ui,u2,u3, instead
, we will write [rir2r3]. Unfortunately, there are no general formulas for the numbers owever, [3] outlines a method for calculating some numbers [rir2r3]. Let u, v,w be vertices of the graph r, W = d(u, v), U = d(v,w), and let V = d(u, w). Since there is exactly one vertex x = u such that d(x,u) = 0, then the number [0jh] is 0 or 1. Hence [0jh] = ¿jw¿hV. Similarly, [i0h] = ¿¿w¿hU and [ij0] = ¿¿u ¿jV.
Another set of equations can be obtained by fixing the distance between two vertices from {u, v,w} and counting the number of vertices located at all possible distances from the third:
5>'h]= j - [0jh] i d
J>h] = - [iOh]
i d
E[iji]= - [ijo]
(3.1)
However, some triplets disappear. For |i — j| > W or i + j < W, we have pW = 0; therefore, [ijh] = 0 for all h € {0, ...,d}. We set
v, w) = ^ QriQsjQth
r,s,t=0
uvw rst -
If the Krein parameter qij = 0, then Sijh(u, v,w) = 0.
We fix vertices u,v,w of a distance-regular graph r of diameter 3 and set
{ijh} =
uvw
ijh
Calculating the numbers
[ijh] =
uvw ijh , [ijh]' = uwv ihj , [ijh]* = vuw _ jih J ' r. -7 wvu [ijh] = hji
uwv ihj , [ijh]* = vuw jih , [ijhr = wvu hji
[ijh]' =
(symmetrization of the triple intersection numbers) can give new relations that make it possible to prove the nonexistence of a graph.
d
4. Graphs with intersection arrays {315, 256, 64; 1,16, 252} and
{1995,1600,320;1,80,1596}
Let r be a distance-regular graph with intersection array {315,256,64; 1,16,252}. By [2, Theorem 4.4.3], the eigenvalues of the local subgraph of the graph r are contained in the interval [—5, 59/5). Since the Terwilliger polynomial (see [4]) is —4(5x — 59)(x + 5)(x + 1)(x — 43), then these eigenvalues lie in [—5, —1] U (59/5.43]. Hence, all nonprinciple eigenvalues are negative and the
local subgraph is a union of isolated (ai + 1)-cliques, a contradiction with the fact that a1 + 1 = 49 does not divide k = 315.
Thus, a distance-regular graph with intersection array {315,256,64; 1,16,252} does not exist.
Let r be a distance-regular graph with intersection array {1995,1600,320; 1,80,1596}. Then r has 1 + 1995 + 39900 + 8000 = 49896 vertices, spectrum 19951, 399495, 1523275, -2126125, and the dual matrix of eigenvalues
26125 -275
55. -209
The Terwilliger polynomial of the graph r is —20(x + 5)(x + 1)(x — 79)(x — 299); hence, the eigenvalues of the local subgraph are contained in [—5, —1] U {79} U {394}.
Note that the multiplicity m1 = 495 of the eigenvalue = 399 is less than k. By the corollary to Theorem 4.4.4 from [2] for b = 61/(01 + 1) = 4, the graph £ = [u] has an eigenvalue —1 — b = —5 of multiplicity at least k — m1 = 1500.
Let the number of eigenvalues 79 of the graph £ be equal to y. Then the sum of eigenvalues of the graph £ is at most —7500 — (494 — y) + 79y + 394; therefore, y > 95. Now twice the number of edges in £ is equal to
786030 = 1995 ■ 394 = ^ mi6>2
i
but not less than
25 ■ 1500 + 399 + 95 ■ 792 + 3942 = 786030.
Hence, £ has spectrum 3941.7995, —1399, — 51500.
Now the number t = ksAs/2 of triangles in £ containing this vertex is equal to ^i mi03/(2v). Therefore,
t = Y1 mA3/(2v) = (3943 + 793 ■ 95 — 399 — 125 ■ 1500)/3990 = 27021
i
and As = 54042/394 is approximately equal to 137.16, a contradiction.
Thus, a distance-regular graph with intersection array {1995,1600,320; 1,80,1596} does not exist.
Theorem 2 is proved.
5. Graph with array {420, 340, 80; 1, 20, 336}
Let r be a distance-regular graph with intersection array {420,340,80; 1,20,336}. Then r is a formally self-dual graph having 1 + 420 + 7140 + 1700 = 9261 vertices, spectrum 4201, 84420 , 07140, —211700, and the dual matrix of eigenvalues
1700 \
—85 .
20 . —64 J
The Terwilliger polynomial of the graph r is —20(x+5)(x+1)(x—16)(x —59) and the eigenvalues of the local subgraph are contained in [—5, —1] U {16} U {79}. If the nonprinciple eigenvalues of a local subgraph are negative, then this subgraph is a union of isolated (a1 + 1)-cliques, a contradiction with the fact that a1 + 1 = 80 does not divide k = 420. Hence, the local subgraph has eigenvalue 6.
Q =
1 495 23275
1 99 175
1 0 —56
1 —99/4 931/4
Q =
1 420 7140
1 84 0
1 0 —21
1 —21 84
Lemma 3. Intersection numbers of a graph r satisfy the equalities
(1) pii = 79, p2i = 340, p32 = 1360, P22 = 5440, p33 = 340,
(2) p2i = 20, p22 = 320, p23 = 80, p22 = 5519, p23 = 1300, p23 = 320;
(3) p?2 = 336, p?3 = 84, p22 = 5460, p33 = 1344, p33 = 271.
Proof. Direct calculations. □
Let u, v, and w be vertices of a graph r, [rst] = [UVT], ^ = r3(u), and let E = Q2. Then E is a regular graph of degree 1344 on 1700 vertices.
Lemma 4. Let d(u, v) = d(u, w) = 3 and d(v,w) = 1. Then the following equalities hold:
(1) [122] = 2re/5 — 136, [123] = [132] = —2re/5 + 472, [133] = 2re/5 — 388;
(2) [211] = re/10 — 38, [212] = [221] = —re/10 + 374, [222] = —14r6/10 + 5576, [223] = [232] = 3re/2 — 490, [233] = —3re/2 + 1834;
(3) [311] = —re/10 + 117, [312] = [321]=re/10 — 34, [322] = re, [323] = [332] = —11re/10 + 1378, [333] = 11re/10 — 1107,
where re € {1010,1020,... , 1170}.
Proof. A simplification of formulas (3.1) taking into account the equalities Sii3(u,v,w) = Si3i(u, v,w) = S3ii(u, v,w) =0. □
By Lemma 4, we have 1010 < [322] = re < 1170.
Lemma 5. Let d(u, v) = d(u, w) = d(v,w) = 3. Then the following equalities hold:
(1) [122] = — ri7 + 336, [123] = [132] = ri7, [133] = — ri7 + 84;
(2) [213] = [231] = ri7, [212] = [221] = — ri7 + 336, [222] = 39ri7/4 + 3444, [223] = [232] = —35ri7/4 + 1680, [233] = 31ri7/4 — 336;
(3) [313] = [331] = — ri7 + 84, [312] = [321] = ri7, [322] = —35ri7/4 + 1680, [323] = [332] = 31ri7/4 — 336, [333] = —27ri7/4 + 522,
where ri7 € {44, 48,... , 76}.
Proof. A simplification of formulas (3.1) taking into account the equalities Sii3(u,v,w) = Si3i(u, v,w) = S3ii(u, v,w) =0. □
By Lemma 5, we have 1015 < [322] = —35ri7/4 + 1680 < 1295.
The number d of edges between E(w) and E — ({w} U A(w)) satisfies the inequalities
359905 = 84 ■ 1010 + 271 ■ 1015 < d < 84 ■ 1170 + 271 ■ 1295 = 449225, 267.786 < 1343 — A < 334.245, 1008.755 < A < 1075.214,
where A is the mean value of the parameter A(E).
Lemma 6. Let d(u, v) = d(u, w) = 3 and d(v,w) = 2. Then the following equalities hold:
(1) [122] = (—64r15 + 4r16 + 7364)/27, [123] = [132] = (64r15 — 4r16 + 1708)/27, [133] = (—64r15 + 4r16 + 560)/27;
(2) [211] = — r15+20, [212] = [221] = (71r15+4r16+6392)/27, [222] = (—17r15 — 13r16+38311)/9, [223] = [232] = (—20r15 + 35r16 + 26095)/27, [233] = (64r15 — 31r16 + 8053)/27;
(3) [311] = r15, [312] = [321] = (—71r15 — 4r16 + 2248)/27, [313] = (44r15 + 4r16 + 20)/27, [322] = (115r15 + 35r16 + 26716)/27, [323] = [332] = (—44r15 — 31r16 + 7297)/27, [333] = r16,
where — 10r15 + 4r16 + 20 is a multiple of 27, r15 € {0,1,..., 20}, and r16 € {0,1,..., 235}.
Proof. A simplification of formulas (3.1) taking into account the equalities S113(u,v,w) = S131(u, v,w) = S311(u, v,w) =0. □
By Lemma 6, we have
998 < [322] = (115r15 + 35r16 + 26716)/27 < 1294. Let us count the number h of pairs of vertices y and z at distance 3 in the graph Q, where
On the one hand, by Lemma 4, we have [323] = -11r6/10 + 1378, where r6 € {1010,1020,..., 1170},
7644 = 8491 < h < 84267 = 22428.
On the other hand, by Lemma 6, we have [313] = (44r15 + 4r16 + 20)/27, where r15 € {0,1,..., 20}, r16 € {0,1,..., 235}, therefore
7644 < Y^(44ri5 + 4ri6) + 995.55 < 22428,
therefore
6648.44 < Y^(44ri5 + 4ri6) < 21432.45,
i
4.946 < Y(11rl5 + ri6)/1344 < 15.947.
If ri 5 = 0, then ri 6 + 5 is a multiple of 27 and ri6 = 22.49,.... If ri 5 = 1, then 2ri 6 + 5 is a multiple of 27 and ri 6 = 11.38,.... In any case,
E(11ri5 + ri6)/1344 > 22,
a contradiction.
Theorem 3 is proved.
□
Conclusion
The following are the main steps in creating a theory of Shilla graphs:
(1) finding a list of feasible intersection arrays of Shilla graphs with b = 6;
(2) classification of Q-polynomial Shilla graphs with b2 = c2.
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